I'm currently working through John M. Lee's Introduction to Topological Manifolds, and I stumbled upon this problem stated (non-verbatim)
Let $Y$ be an uncountable well-ordered set such that for each $x \in Y$ there are at most countably many $y \in Y$ with $y < x$. Equip $R := Y \times [0, 1)$ with the dictionary order. Together with the order topology, $R$ is called a long ray. If $L := R \vee R$ is the wedge sum obtained by identifying both $(y_{0}, 0)$, then $L$ is called a long line, where $y_{0}$ is the smallest element of $Y$. Prove that $L$ is locally Euclidean, Hausdorff, and first countable but not second countable.
I'm attempting to prove that $L$ is locally Euclidean. Here is the sketch of my attempt: I took an element $(y, r) \in R$ such that $(y, r) \neq (y_{0}, 0)$. Then either $y \neq y_{0}$ or $r \neq 0$. If $r \neq 0$, then $0 < r < 1$. Hence $(y, 0) \in I := ((y, 0), (y, (r + 1) / 2))$. But if $y \neq y_{0}$, then $y_{0} < y$. But since $Y$ is uncountable, then there exists $x_{0} \in Y$ such that $y < x_{0}$, and so $(y, r) \in J := ((y_{0}, 0), (x_{0}, 0))$.
$I$ and $J$ are open subsets of $R$, and my hypothesis are (1) $I$ is homeomorphic to the open interval $(0, (r + 1) / 2)$ and (2) $J$ is homeomorphic to the set of all positive real numbers, given that there are infinitely many $x \in Y$ such that $x < y$. I think it is easy to prove (1), but I'm not sure how to prove (2). I attempted to arrange the elements of $X := \{ x \in Y : x < y \}$ so that $X = \{ y_{0} < y_{1} < \dotsc y < \dotsc \}$ using the well-ordering of $Y$. But since $X$ could be infinite, I'm not sure if I can do that. But assuming that the attempt was a success, I will define a map $f$ from $X$ to the set $\mathbb{Z}_{\geq 0}$ of all nonnegative integers by $f(n) = y_{n}$ so that $f$ is a bijective monotone function, and I will extend $f$ to a homeomorphism $f : J \to \mathbb{R}_{> 0}$.
A little help would be appreciated.
You in general can't find monotonic bijection from $X$ to $\mathbb N$ (even for $X = \omega + 1$). But you don't actually need to arrange all elements of $X$ into a sequence, you just need to find a monotonic sequence of elements of $X$ that converges to $y$.
Assume $X = \{x_i | i \in \mathbb N\}$ (it's possible, as $X$ is countable). Now, let $y_1 = x_1$, and $y_{n + 1} = \max(y_n, x_{n + 1}) + 1$ (here and below $y + 1$ means element following $y$, which exists because $Y$ is well-ordered).
To prove the statement, we can prove by (transfinite) induction that $[(x, 0), (y, 0))$ if homeomorphic to $[0, 1)$ for any $y > x \geq y_0$.
Case $y = y' + 1$ is trivial.
If $y$ is limit, then there is countable monotonic sequence $y_i < y$ s.t. $y = \sup_i y_i$. Wlog assume $y_1 = x$.
Now, we have $[(x, 0), (y, 0)) = \sqcup_i [(y_i, 0), (y_{i + 1}, 0))$. By induction assumption, each $[(y_i, 0), (y_{i + 1}, 0))$ is homeomorphic to $[i, i + 1)$, and by combining these homeomorphisms we get homeomorphism $[(x, 0), (y, 0)) \to [0, +\infty)$, which is homeomorphic to $[0, 1)$.